记cn=anbn,求数列{cn}的前n项和Tn
发布网友
发布时间：2023-08-23 09:48
共1个回答
热心网友
时间：2024-11-26 04:31
Sn =n^2+2n
an = Sn -S(n-1)
= 2n+1
bn = b1q^(n-1)
b1 = a1/2 = 3/2
b3(a3-a1) = b1
(3/2)q^2 [ 7-3] = 3/2
q^2 =1/4
q= 1/2
bn = 3.(1/2)^n
(2)
cn =anbn
= (2n+1) ( 3. (1/2)^n)
= 3 [n. (1/2)^(n-1)] + 3(1/2)^n
Tn = c1+c2+..+cn
= 3{ summation(i:1->n) i.(1/2)^(i-1) } + 3( 1- (1/2)^n)
consider
1+x+x^2+...+x^n = (x^(n+1) -1)/(x-1)
1+2x+...+nx^(n-1) = [(x^(n+1) -1)/(x-1)]'
= [ nx^(n+1) -(n+1)x^n +1]/(x-1)^2
put x= 1/2
{ summation(i:1->n) i.(1/2)^(i-1) }
= 4[ n(1/2)^(n+1) -(n+1)(1/2)^n +1]
= 4 -2(n+2)(1/2)^n
Tn = 3{ summation(i:1->n) i.(1/2)^(i-1) } + 3( 1- (1/2)^n)
= 3{4 -2(n+2)(1/2)^n} +3( 1- (1/2)^n)
= 15 - (2n+7)(1/2)^n
